Chứng minh:
\(\dfrac{1}{3^2}+\dfrac{1}{6^2}+\dfrac{1}{9^2}+...+\dfrac{1}{2003^2}< \dfrac{1}{5}\)
NN
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Tính giá trị của các biểu thức sau 1) A1+2+2^2+...+2^{2015} 2) Bleft(dfrac{1}{4}-1right)cdotleft(dfrac{1}{9}-1right)cdotleft(dfrac{1}{16}-1right)cdotcdotcdotcdotcdotleft(dfrac{1}{400}-1right) 3) Cleft(dfrac{1}{4cdot9}+dfrac{1}{9cdot14}+dfrac{1}{14cdot19}+...+dfrac{1}{44cdot49}right)cdotdfrac{1-3-5-7-...-49}{89} 4) Ddfrac{2^{12}cdot3^5-4^6cdot9^2}{left(2^2cdot3right)^6+8^4cdot3^5}-dfrac{5^{10}cdot7^3-25^5cdot49^2}{left(125cdot7right)^3+5^9cdot14^3} 5) Edfrac{dfrac{1}{2003}+dfrac{1}{2004}-dfrac{1}...
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Tính giá trị của các biểu thức sau 1) \(A=1+2+2^2+...+2^{2015}\) 2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\) 3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\) 4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\) 5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\) 6) Cho 13+23+...+103=3025 Tính S= 23+43+63+...+203
Bài 1:
a, Tính \(\dfrac{5.4^{15}.9^{9} - 4.3^{20}.8^{9}}{5.2^{9}.6^{19}-7.2^{29}.27^{6}}\)
b,Tìm x biết:
\(1\dfrac{1}{30}:(24\dfrac{1}{6}-24\dfrac{1}{5}) -\dfrac{1\dfrac{1}{2}-\dfrac{3}{4}}{4x-\dfrac{1}{2}}=(-1\dfrac{1}{15}):(8\dfrac{1}{5}-8\dfrac{1}{3})\)
Bài 2: Chứng minh số:
222...22200333...333(2001c/s 2; 2003 c/s3)
Tính giá trị của các biểu thức sau1) A1+2+2^2+...+2^{2015}2) Bleft(dfrac{1}{4}-1right)cdotleft(dfrac{1}{9}-1right)cdotleft(dfrac{1}{16}-1right)cdotcdotcdotcdotcdotleft(dfrac{1}{400}-1right)3) Cleft(dfrac{1}{4cdot9}+dfrac{1}{9cdot14}+dfrac{1}{14cdot19}+...+dfrac{1}{44cdot49}right)cdotdfrac{1-3-5-7-...-49}{89}4) Ddfrac{2^{12}cdot3^5-4^6cdot9^2}{left(2^2cdot3right)^6+8^4cdot3^5}-dfrac{5^{10}cdot7^3-25^5cdot49^2}{left(125cdot7right)^3+5^9cdot14^3}5) Edfrac{dfrac{1}{2003}+dfrac{1}{2004}-dfrac{1}{2005...
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Tính giá trị của các biểu thức sau
1) \(A=1+2+2^2+...+2^{2015}\)
2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\)
3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
6) Cho 13+23+...+103=3025
Tính S= 23+43+63+...+203
Tính giá trị của các biểu thức sau
1) A1+2+2^2+...+2^{2015}
2) Bleft(dfrac{1}{4}-1right)cdotleft(dfrac{1}{9}-1right)cdotleft(dfrac{1}{16}-1right)cdotcdotcdotcdotcdotleft(dfrac{1}{400}-1right)
3) Cleft(dfrac{1}{4cdot9}+dfrac{1}{9cdot14}+dfrac{1}{14cdot19}+...+dfrac{1}{44cdot49}right)cdotdfrac{1-3-5-7-...-49}{89}
4) Ddfrac{2^{12}cdot3^5-4^6cdot9^2}{left(2^2cdot3right)^6+8^4cdot3^5}-dfrac{5^{10}cdot7^3-25^5cdot49^2}{left(125cdot7right)^3+5^9cdot14^3}
5) Edfrac{dfrac{1}{2003}+dfrac{1}{2004}-d...
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Tính giá trị của các biểu thức sau
1) \(A=1+2+2^2+...+2^{2015}\)
2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\)
3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
6) Cho 13+23+...+103=3025
Tính S= 23+43+63+...+203
1) \(A=1+2+2^2+2^3+......+2^{2015}\)
\(\Leftrightarrow2A=2+2^2+2^3+......+2^{2016}\)
\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{2016}\right)-\left(1+2+2^2+2^3+......+2^{2015}\right)\)
\(\Leftrightarrow A=2^{2016}-1\)
Vậy \(A=2^{2016}-1\)
6)Ta có: \(13+23+33+43+.......+103=3025\)
\(\Leftrightarrow2.13+2.23+2.33+2.43+.......+2.103=2.3025\)
\(\Leftrightarrow26+46+66+86+.......+206=6050\)
\(\Leftrightarrow\left(23+3\right)+\left(43+3\right)+\left(63+3\right)+\left(83+3\right)+.......+\left(203+3\right)=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+3.10=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+=6050-30\)
\(\Leftrightarrow23+43+63+83+.......+203+=6020\)
Vậy S=6020
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b, B có 19 thừa số
=> \(-B=(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})...(1-\frac{1}{400}) \)
<=>\(-B=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(20-1)(20+1)}{4.9.16...400} \)
<=>\(-B=\frac{(1.2.3.4...19)(3.4.5...21)}{(2.3.4.5.6...20)(2.3.4.5...20)} \)
<=>\(-B=\frac{21}{20.2} =\frac{21}{40} \)
<=>\(B=\frac{-21}{40} \)
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Cho biểu thức A=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+\(\dfrac{1}{7^2}\)+\(\dfrac{1}{8^2}\)+\(\dfrac{1}{9^2}\)+\(\dfrac{1}{10^2}\)
Chứng minh rằng A<1
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2\cdot2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3\cdot3}< \dfrac{1}{2\cdot3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4\cdot4}< \dfrac{1}{3\cdot4}\)
...
\(\dfrac{1}{9^2}=\dfrac{1}{9\cdot9}< \dfrac{1}{8\cdot9}\)
\(\dfrac{1}{10^2}=\dfrac{1}{10\cdot10}< \dfrac{1}{9\cdot10}\)
\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow A< 1-\dfrac{1}{10}\)
\(\Rightarrow A< \dfrac{9}{10}\)
\(\Rightarrow A< 1\) (vì: \(\dfrac{9}{10}< 1\))
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H24
24 tháng 11 2023 lúc 12:15
cho \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\) chứng minh \(\dfrac{2}{5}< A< \dfrac{8}{9}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}\)
=>\(A< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{8\cdot9}\)
=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}=1-\dfrac{1}{9}=\dfrac{8}{9}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}\)
=>\(A>\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
=>\(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
=>\(A>\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{5}{10}-\dfrac{1}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)
Do đó: \(\dfrac{2}{5}< A< \dfrac{8}{9}\)
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Chứng minh \(\dfrac{1}{5}\)< \(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+\(\dfrac{1}{7^2}\)+......+\(\dfrac{1}{99^2}\)+\(\dfrac{1}{100^2}\)<\(\dfrac{1}{3}\)
\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{100\cdot101}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{4}-\dfrac{1}{101}>\dfrac{1}{4}-\dfrac{1}{20}=\dfrac{1}{5}\left(1\right)\)
\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\left(2\right)\) Từ (1) và (2) \(\Rightarrow\dfrac{1}{5}< \dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}< \dfrac{1}{3}\)
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Tính:
a) (6 : dfrac{3}{5} 1dfrac{1}{6} x dfrac{6}{7} ) : ( 4dfrac{1}{5} x dfrac{10}{11} + 5dfrac{2}{11} )
b) (1-dfrac{1}{2}) x (1-dfrac{1}{3}) x (1-dfrac{1}{4}) x ..... x (1-dfrac{1}{2003}) x (1-dfrac{1}{2007})
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Tính:
a) (6 : \(\dfrac{3}{5}\) \(1\dfrac{1}{6}\) x \(\dfrac{6}{7}\) ) : ( \(4\dfrac{1}{5}\) x \(\dfrac{10}{11}\) + \(5\dfrac{2}{11}\) )
b) (\(1-\dfrac{1}{2}\)) x (\(1-\dfrac{1}{3}\)) x (\(1-\dfrac{1}{4}\)) x ..... x (\(1-\dfrac{1}{2003}\)) x (\(1-\dfrac{1}{2007}\))
Bài 1:Chứng tỏ rằng:Bdfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+dfrac{1}{5^2}+dfrac{1}{6^2}+dfrac{1}{7^2}dfrac{1}{8^2}1Bài 2:Chứng tỏ rằng:Edfrac{3}{4}+dfrac{8}{9}+dfrac{15}{16}+...+dfrac{2499}{2500}1Bài 3:Chứng tỏ rằng:1dfrac{2011}{2020^2+1}+dfrac{2021}{2020^2+2}+dfrac{2021}{2020^3+3}+...+dfrac{2021}{2020^3+2020} 2
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Bài 1:Chứng tỏ rằng:B=\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{5^2}\)+\(\dfrac{1}{6^2}\)+\(\dfrac{1}{7^2}\)\(\dfrac{1}{8^2}\)<1
Bài 2:Chứng tỏ rằng:E=\(\dfrac{3}{4}\)+\(\dfrac{8}{9}\)+\(\dfrac{15}{16}\)+...+\(\dfrac{2499}{2500}\)<1
Bài 3:Chứng tỏ rằng:1<\(\dfrac{2011}{2020^2+1}\)+\(\dfrac{2021}{2020^2+2}\)+\(\dfrac{2021}{2020^3+3}\)+...+\(\dfrac{2021}{2020^3+2020}\)< 2
1:
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)
...
\(\dfrac{1}{8^2}< \dfrac{1}{7\cdot8}\)
=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+..+\dfrac{1}{7\cdot8}\)
=>\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}=\dfrac{7}{8}< 1\)
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